Question

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106 m).

Answer 1

◆ Answer -

F = 9.77 N

● Explanation -

# Given -

M = 6×10^24 kg

m = 1 kg

R = 6.4×10^6 m

# Solution -

Gravitational force between earth and the object is given by -

F = GMm/R²

F = 6.67×10^-11 × 6x10^24 × 1 / (6.4x10^6)²

F = 9.77 N

Therefore, gravitational force between the earth and an object is 9.77 N.

Hope this helps you...

Answer 2

Answer:

9.8N

Explanation:

Given :

• Mass of the earth, M => 6 × 1024 kg
• Mass of the body, m => 1 kg
• Radius of the earth,r => 6.4 × 106
• Unversal gravitational constant,G =>6.7 × 10−11 Nm2 kg−2

To Find :

• Magnitude of the gravitational force between the earth

Solution :

We know that:-

$\sf{}F=\dfrac{GMm}{r^2}$

Substitue their values and solve it

$\Rightarrow \sf{}\dfrac{6.7\times 10^{-11}\times6\times^{24}\times1}{(6.4\times10^6)^2}$

$\sf{}\Rightarrow 9.8N$

This shows that the earth exerts a force of 9.8N on the body of a mass of 1kg and the body will exert an equal force of attraction of 9.8N on the earth.